Kitonum

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These are answers submitted by Kitonum

restart;

Pbc:=proc(n::posint)
local Ind, Var1, Var2;
uses combinat;
Ind:=permute([0$n,1$n],n):
Var1:=[seq(x||i,i=0..n-1)]: Var2:=[seq(y||i,i=0..n-1)]:
{seq(seq(inner(Var1,ind1)*inner(Var2,ind2), ind1=Ind), ind2=Ind)};
end proc:

Examples of use (for n = 2 the result coincides with Carl's one):
Pbc(1);
Pbc(2);
Pbc(3);
             

There will be infinitely many such vectors. All of them belong to a plane perpendicular to the original vector. Therefore, the set of all solutions depends on two parameters:

restart;
PV:=(a,b,c)->solve(a*x+b*y+c*z=0,{x,y,z}):

# Examples of use:
V:=<1,2,3>;
PV(V[1],V[2],V[3]);
V1:=eval(<x,y,z>, %); # General solution
V.V1; # Check
V2:=eval(V1,[y=1,z=1]);  # One of the solutions

                            

Define functions as procedures, not expressions:

restart;
h:=x->1/(1+exp(-x));
hh:=(w,x,b)->ln(h(w*x + b));
diff(hh(w,x,b), w);  # Or  D[1](hh)(w,x,b);
normal(expand(%));

                         


I always refer to previous objects using their names or by  %  and so on.

Edit.

Here is a more traditional way:

restart;
IntegrationTools:-Change(int(arccos(x)*arcsin(x),x), x=sin(t));
eval(%, sin(t)=x);

          

 

 


 

restart;
b:=proc(n)
option remember;
  if n=0 then return 0 elif n=1 then return 1
  else
   (b(n-1)+b(n-2))/2:
  fi;
end proc:

# Examples of use: 

S:=seq(b(n),n=0..10);
plot([$ 0..10],[S]);

                


You can obtain an explicit formula for the nth member by  rsolve  command:

restart;
# The formula for nth term
rsolve({b(n)=(b(n-1)+b(n-2))/2,b(0)=0,b(1)=1}, b(n));

                       

seq1.mw
 

This is certainly a bug. Do

with(geometry):
point(o, 0, 0);
point(A, 0, 1);
point(d, 0, 2);
point(F, 0.8944271920, 1.4472135960);
line(lOD, [o, d]);
line(lAF, [A, F]);
alpha1 := FindAngle(lOD, lAF);
alpha:=min(alpha1,Pi-alpha1);

                                                alpha := 1.107148718

f:=y->sqrt((1+y)/2):
cos(Pi/2)=0:
cos(Pi/2^n)=(f@@(n-1))(0):

# Examples
cos(Pi/16)=(f@@3)(0);
cos(Pi/64)=(f@@5)(0);

                     

 

Any command, if you add the  %  symbol in front of it, becomes inert. If we replace  Sum  with  %sum , then the bug disappears. See below

 

Download simplify_new.mw

Your system contains too many parameters  Q11A, Q11Ay, Q11Az, Q11Iy, Q11Iz, Q11J, Q16A, Q16Ay, Q16Az, Q16Iz, Q16J, Q55A, Q55Ay, Q55Iy, Q66A, Q66Az, Q66Iz  (total 17 ones) with unknown values. If you specify the values of these parameters, Maple easily solves this system. I took parameter values from 1 to 17.

 

Download copy1_new.mw

Here is a procedure for this:

LeastDegree:=proc(P::polynom, var::{set,list})
local L;
L:=sort(map(p->degree(p,var),[op(P)]));
select(p->degree(p,var)=L[1],P);
end proc:


Examples of use:

f:= 100*x^2*y^2 + 35*x^2*y + 45*y:
g:= 13*x^2*y^2 + x*a*y^2 + 2*y*x^2:
LeastDegree(f, [x,y]);
LeastDegree(g, [x,y]);
                                            
45*y
                                     a*x*y^2+2*x^2*y

Edit.

Specify the range for x:

plot(sin(4*x)+1/3*cos(6*x), x=0..2*Pi);

 

This animation shows thin cylindrical disks, summing up the volumes of which and passing to the limit, we get the exact volume:

restart;
f:=x^(1/2):
g:=x^2/8:
X:=r*cos(phi): Y:=r*sin(phi):
P:=plot3d(eval([[X,Y,f],[X,Y,g]],x=r), r=0..4, phi=0..2*Pi, style=surface, color=["Khaki","LightBlue"], scaling=constrained, axes=normal, labels=[z,x,y], orientation=[20,80], transparency=0.5):
F:=y->plots:-display(plot3d([[r*cos(phi),r*sin(phi),y],[r*cos(phi),r*sin(phi),y+h]], r=y^2..sqrt(8*y), phi=0..2*Pi, style=surface, color=gold), plot3d([[y^2*cos(phi),y^2*sin(phi),H],[sqrt(8*y)*cos(phi),sqrt(8*y)*sin(phi),H]], H=y..y+h, phi=0..2*Pi, style=surface, color=gold)):
h:=0.15:
plots:-animate(F,[y], y=0..2-h, frames=60, background=P);

                   

# Volume of this body in 2 ways
Int(Pi*(8*y-y^4), y=0..2)=int(Pi*(8*y-y^4), y=0..2);  # Washer (disk) method
Int((sqrt(x)-x^2/8)*2*Pi*x, x=0..4)=int((sqrt(x)-x^2/8)*2*Pi*x, x=0..4);  # Shell (cylinder) method
                      

 
 

I guess that you consider this function to be real-valued, so its domain will be x>0 and x<1 . The solution to your problem is based on 2 key properties that are easy to prove:
1. The graph of the function is symmetric with respect to the line  x = 1/2 .
2. For any  a>2 the function  f(x)  has  1 local  maximum at  x=1/2  and 2 local minimums  x1  and  x2 

By virtue of symmetry, all these conditions (df/dx(x=x1)=df/dx(x=x2)=0 and f(x1)=f(x2)) are satisfied for these points x1  and  x2. Below the calculation of  x1  and  x2  and the plotting for  a=3 .

restart;
f:=(x,a)->a*x*(1-x)+x*ln(x)+(1-x)*ln(1-x);
limit(f(x,a),x=0);
plot(f(x,3),x=0..1);
x1:=fsolve(diff(f(x,3),x));
x2:=1-x1;
plot(f(x,3), x=x1..x2);

                    

 

                           

 

A:=d*n0/dt+d*epsilon*n1/dt+d*epsilon^(3/2)*n2/dt+d*epsilon^2*n3/dt+epsilon*d*n0*u1/dx+epsilon^(3/2)*d*n0*u2/dx+d*n0*epsilon^2*u3/dx+d*epsilon^2*n1*u1/dx+d*epsilon^(5/2)*n1*u2/dx:
coeff(A,epsilon^(3/2));

                                     d*n2/dt+d*n0*u2/dx        

The list  L2 contains all the options for representing numbers from your list in a sum of no more than 5 different squares.

For some reason, the document is not displayed inline. Therefore, I copied the code and the output fragment:

restart;
Squares:=[seq(i^2,i=0..12)]:
k:=0:
for n from 129 to 129+13^2 do
for m from 1 to 5 do
for c in combinat:-choose(Squares,m) do
if n=add(c) then k:=k+1; L[k]:=[n,c] fi;
od: od: od:
L:=convert(L,list):
L1:={seq(p[1]=`+`(seq(sqrt(p[2][i])^`2`,i=1..nops(p[2]))),p=L)}:
L2:=[ListTools:-Categorize((x,y)->lhs(x)=lhs(y),[L1[]])]:
for p in L2 do
lhs(p[1])=`or`(seq(rhs(p[k]),k=1..nops(p)));
od;

  



Download sums_squares2.mw

Edit.

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